That’s the amount that we shifted our wave, so it is also correct! Formulas So Far We want to take atan2(sine, cosine), aka atan2(-0.25, 0.97) which gives us a result of -0.25 radians. To get the starting angle (phase) of the wave, we use inverse tangent of sine over cosine. That is correct! Our wave’s amplitude is 1.0. To get the amplitude of our phase shifted wave, we treat those values as a vector, and get the magnitude. We took a cosine wave and shifted it over a little bit so our combined answer is that it’s mostly a cosine wave, but is heading towards being a sine wave (technically a sine wave that has an amplitude of -1, so is a negative sine wave, but is still a sine wave). Those numbers kind of makes sense if you think about it. Since we know both of those numbers are twice the actual amplitude, we can divide them both by two to get: The solution to this problem is actually to multiply the data points by both cosine and sine waves, and keep both results.įor this example We’ll take a cosine wave and shift it 0.25 radians to give us samples: 0.97, -0.25, -0.97, 0.25. We might even have some other type of wave that is neither one! That’s not very useful if we have to know whether we are looking for a sine or cosine wave though. We get results consistent with before, where our answer is the amplitude times two. In this case, if instead of multiplying by cosine, we multiply by sine, we get what we expect: 0*0 + 1*1 + 0*0 + -1*-1 = 2. When we multiply those values by our cosine wave values we get: 0*1 + 1*0 + 0*-1 + -1*0 = 0. What happens if our stream of data has something other than a cosine wave in it though? In general, you will need to divide by N / 2 where N is the number of samples you have. The actual amplitude is 5, so we can see that our result was still twice the amplitude. To show that this works for any amplitude, here’s the same 1hz cosine wave data with an amplitude of five: 5, 0, -5, 0. We can divide by two to get the real amplitude. The result we get is 2, which is twice the amplitude of the data points we had.

We could then multiply each point by a corresponding point on a cosine wave, add sum them together: One way would be to make a cosine wave of that frequency and multiply each sample of your wave by the corresponding samples.įor instance, let’s say that we have a data stream of four values that represent a 1hz cosine wave: 1, 0, -1, 0. If you had a stream of data that had a single frequency in it at a constant amplitude, and you wanted to know the amplitude of that frequency, how could you figure that out? This is useful if you want to analyze data, but can also be useful if you want to modify the frequencies then use the inverse discrete Fourier transform to generate the frequency modified data. The discrete Fourier transform takes in data and gives out the frequencies that the data contains. I’ve figured out some things which have really helped my intuition, and made it a lot simpler in my head, so I wanted to write these up for the benefit of other folks, as well as for my future self when I need a refresher. I’ve been working on getting a better understanding of the Discrete Fourier Transform.

Inverse fourier transform calculator